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我有一个我真的不知道如何开始的查询。我希望有人能帮我解决这个问题。我将从解释表格开始

我有一个包含四列的设备表：Device_Id、Device_Status、Begin_dt、End_dt 有 6 种不同的状态，其中 3（为简单起见，可以说状态 1、4 和 5）表示设备“在线”

该表的一个示例可能是

Id | Status| Begin                   |  End
001|     1 | 2012-09-01 00:00:00.000 | 2012-09-01 01:00:00.000
001|     2 | 2012-09-01 01:00:00.000 | 2012-09-01 01:35:00.000
001|     1 | 2012-09-01 01:35:00.000 | 2012-09-01 02:05:00.000
003|     1 | 2012-09-01 05:00:00.000 | 2012-09-01 07:02:00.000
004|     1 | 2012-09-01 01:00:00.000 | 2012-09-01 01:35:00.000
003|     2 | 2012-09-01 07:02:00.000 | NULL

我的查询需要返回 TIME 的总和，所有设备的状态都表明 24 小时内每个小时都处于“在线”状态。

所以我的回报应该看起来像

Hour| Online_Time
0   | 5:30:12.11
1   | 3:30:12.11
2   | 4:30:12.11
3   | 5:30:12.11
4   | 6:30:12.11
5   | 4:00:00.00
6   | 1:30:12.11
7   | 3:30:12.11
8   | 4:30:12.11
etc |

因此，对于一天中的每个小时，我可以有超过 1 小时的在线时间（显然），因为例如，如果我有 5 台设备在整个小时内全部在线，那么我将有 5 小时的在线时间。

这有点复杂，我希望我能很好地解释它，非常感谢任何可以提供帮助或提供建议的人。

-J

with const as (
    select dateadd(hour, 1, cast(cast(getdate() -1 as date) as datetime)) as midnnight
    ),
allhours as (
    select 0 as hour, midnight as timestart, dateadd(hour, 1, timestart) as timeend from const union all
    select 1 as hour, dateadd(hour, 1, midnight), dateadd(hour, 2, midnight) from const union all
    select 2 as hour, dateadd(hour, 2, midnight), dateadd(hour, 3, midnight) from const union all
    select 3 as hour, dateadd(hour, 3, midnight), dateadd(hour, 4, midnight) from const union all
    select 4 as hour, dateadd(hour, 4, midnight), dateadd(hour, 5, midnight) from const union all
    select 5 as hour, dateadd(hour, 5, midnight), dateadd(hour, 6, midnight) from const union all
    select 6 as hour, dateadd(hour, 6, midnight), dateadd(hour, 7, midnight) from const union all
    select 7 as hour, dateadd(hour, 7, midnight), dateadd(hour, 8, midnight) from const union all
    select 8 as hour, dateadd(hour, 8, midnight), dateadd(hour, 9, midnight) from const union all
    select 9 as hour, dateadd(hour, 9, midnight), dateadd(hour, 10, midnight) from const union all
    select 10 as hour, dateadd(hour, 10, midnight), dateadd(hour, 11, midnight) from const union all
    select 11 as hour, dateadd(hour, 11, midnight), dateadd(hour, 12, midnight) from const union all
    select 12 as hour, dateadd(hour, 12, midnight), dateadd(hour, 13, midnight) from const union all
    select 13 as hour, dateadd(hour, 13, midnight), dateadd(hour, 14, midnight) from const union all
    select 14 as hour, dateadd(hour, 14, midnight), dateadd(hour, 15, midnight) from const union all
    select 15 as hour, dateadd(hour, 15, midnight), dateadd(hour, 16, midnight) from const union all
    select 16 as hour, dateadd(hour, 16, midnight), dateadd(hour, 17, midnight) from const union all
    select 17 as hour, dateadd(hour, 17, midnight), dateadd(hour, 18, midnight) from const union all
    select 18 as hour, dateadd(hour, 18, midnight), dateadd(hour, 19, midnight) from const union all
    select 19 as hour, dateadd(hour, 19, midnight), dateadd(hour, 20, midnight) from const union all
    select 20 as hour, dateadd(hour, 20, midnight), dateadd(hour, 21, midnight) from const union all
    select 21 as hour, dateadd(hour, 21, midnight), dateadd(hour, 22, midnight) from const union all
    select 22 as hour, dateadd(hour, 22, midnight), dateadd(hour, 23, midnight) from const union all
    select 23 as hour, dateadd(hour, 23, midnight), dateadd(hour, 24, midnight) from const union all
   ) 
select ah.hour,
   sum(datediff(ms, (case when ah.timestart >= dt.Begin_Dt then timestart else dt.Begin_Dt end),
                    (case when ah.timeend <= dt.End_Dt then ah.timeend else dt.End_Dt end))) as totalms,
   cast(dateadd(ms, sum(datediff(ms, (case when ah.timestart >= dt.Begin_Dt then timestart else dt.Begin_Dt end),
                                 (case when ah.timeend <= dt.End_Dt then ah.timeend else dt.End_Dt end))),0) as time
                                 ) as totalTime
from allhours as ah left outer join
     dataTable as dt
     on ah.timestart< coalesce(dt.End_dt, getdate()) and
        ah.timeend >= dt.Begin_Dt
group by ah.hour
order by ah.hour

这就是我现在所拥有的，我在“）”上遇到错误

select 23 as hour, dateadd(hour, 23, midnight), dateadd(hour, 24, midnight) from const union all
   )  <----- Incorrect syntax near ')'. Expecting SELECT, or '('.
select ah.hour,
4

2 回答 2

1

您的问题似乎是时间跨度需要以小时为单位。因此，您需要从一天中的所有时间开始。然后计算重叠，总结差异（以毫秒为单位）并将所有内容转换回输出时间。

with const as (
        select dateadd(hour, 1, cast(cast(getdate() -1 as date) as datetime)) as midnight            
    ),
    allhours as (
        select 0 as hour, midnight as timestart, dateadd(hour, 1, midnight) as timeend from const union all
        select 1 as hour, dateadd(hour, 1, midnight), dateadd(hour, 2, midnight) from const union all
        select 2 as hour, dateadd(hour, 2, midnight), dateadd(hour, 3, midnight)  from const union all
        . . .
        select 23 as hour, dateadd(hour, 23, midnight), dateadd(hour, 24, midnight) from const
    )
select ah.hour,
       sum(datediff(ms, (case when ah.timestart >= dt.begin then timestart else dt.begin end),
                        (case when ah.timeend <= dt.end then ah.timeend else dt.end end)
                   ) 
           ) as totalms,
       cast(dateadd(ms, sum(datediff(ms, (case when ah.timestart >= dt.begin then timestart else dt.begin end),
                                     (case when ah.timeend <= dt.end then ah.timeend else dt.end end)
                                    )
                           ),
                     0) as time
           ) as totalTime
from allhours ah left outer join
     DeviceTable dt
     on ah.timestart< coalesce(dt.end, getdate()) and
        ah.timeend >= dt.begin
group by ah.hour
order by ah.hour

此外，要完成这项工作，您需要将“开始”和“结束”用双引号或方括号括起来。这些是 T-SQL 中的保留字。你需要替换“......” 额外的线路从 3 到 22 小时。

于 2012-09-12T13:31:16.830 回答
0

这样的事情可能会让你开始。

select  SUM(datediff(second, Begin_dt, End_dt)) as seconds_online
        ,DATEPART(hour, Begin_dt) as [hour]
from table
where device_status in (1,4,5)
group by DATEPART(hour, Begin_dt)

要格式化结果，您可能需要遵循以下问题：

SQL - 秒到日、时、分、秒

于 2012-09-12T13:03:23.547 回答